Question

find the magnitude of the vector sum \\(\vec{a} + \vec{b}\\).
vector a is 3.60 m long in a \\(45.0^\circ\\) direction.
vector b is 38.4 m long in a \\(162^\circ\\) direction.

Explanation:

Step1: Resolve Vector A into components

To find the x and y components of vector \(\vec{A}\), we use the formulas \(A_x = A\cos\theta_A\) and \(A_y = A\sin\theta_A\), where \(A = 3.60\space m\) and \(\theta_A = 45.0^\circ\).

\(A_x = 3.60\cos(45.0^\circ)\)
\(A_x = 3.60\times\frac{\sqrt{2}}{2} \approx 3.60\times0.7071 \approx 2.5456\space m\)

\(A_y = 3.60\sin(45.0^\circ)\)
\(A_y = 3.60\times\frac{\sqrt{2}}{2} \approx 2.5456\space m\)

Step2: Resolve Vector B into components

For vector \(\vec{B}\), with \(B = 38.4\space m\) and \(\theta_B = 162^\circ\), we use \(B_x = B\cos\theta_B\) and \(B_y = B\sin\theta_B\).

\(B_x = 38.4\cos(162^\circ)\)
\(\cos(162^\circ)=\cos(180^\circ - 18^\circ)=-\cos(18^\circ)\approx - 0.9511\)
\(B_x = 38.4\times(-0.9511)\approx - 36.5222\space m\)

\(B_y = 38.4\sin(162^\circ)\)
\(\sin(162^\circ)=\sin(180^\circ - 18^\circ)=\sin(18^\circ)\approx0.3090\)
\(B_y = 38.4\times0.3090\approx11.8656\space m\)

Step3: Find the components of the resultant vector \(\vec{R}=\vec{A}+\vec{B}\)

The x - component of the resultant vector \(R_x = A_x + B_x\)
\(R_x = 2.5456- 36.5222\approx - 33.9766\space m\)

The y - component of the resultant vector \(R_y = A_y + B_y\)
\(R_y = 2.5456 + 11.8656\approx14.4112\space m\)

Step4: Calculate the magnitude of the resultant vector

The magnitude of the resultant vector \(\vec{R}\) is given by \(R=\sqrt{R_x^2 + R_y^2}\)

\(R=\sqrt{(-33.9766)^2+(14.4112)^2}\)
First, calculate the squares:
\((-33.9766)^2\approx1154.41\)
\((14.4112)^2\approx207.68\)

Then sum them: \(1154.41 + 207.68=1362.09\)

Take the square root: \(R=\sqrt{1362.09}\approx36.91\space m\)

Answer:

The magnitude of the vector sum \(\vec{A}+\vec{B}\) is approximately \(\boldsymbol{36.9\space m}\) (rounded to three significant figures)